@Markiyan Hirnyk  Of course my reasoning is incorrect. Thank you for this error message.

@Markiyan Hirnyk  I meant that, for example, both numbers  F1(x0, y0, z0))  and  F1(x5, y5, z5)  are positive or both ones are negative numbers. In both cases should be  F1(x0, y0, z0))*F1(x5, y5, z5)>0 . Function  sign  we have not use.

Unfortunately, Maple does not solve even the simplest equations containing the function  floor :

solve(floor(x) = 1, x);

                RootOf(floor(_Z)-1)

@Markiyan Hirnyk  If you remove the empty line   [``,``,``,``,``]  then there are no problems.

@Markiyan Hirnyk  In the attached file there is an ordinary Fourier series.

@Markiyan Hirnyk  See the timing in M 12:

restart;
a := rsolve({X(0) = 0, X(1) = 1, X(2) = 2, X(n) = n*(X(n-3)+X(n-2)+X(n-1))}, X, makeproc):

t:=time(): ceil(log[10](a(10000))); time()-t;

                                                                 35664

                                                                 11.578

restart;
NumberOfDigits:=proc(N)
local X, n;
X[0]:=0; X[1]:=1; X[2]:=2;
   for n from 3 to N do
   X[n]:=n*(X[n-3]+X[n-2]+X[n-1]);
   od;
length(%);
end proc:

t:=time(): NumberOfDigits(10000); time()-t;

                                                           35664

                                                           0.297

restart;
X:= proc(n)
option remember;
n*(X(n-3)+X(n-2)+X(n-1))
end proc:

t:=time():  X(0),X(1),X(2):= 0,1,2:
length(X(10000));  time()-t;

                                                            35664

                                                            0.312

I think the reason is that  length  command a lot more effective than ceil@log[10] .

@Carl Love  Unfortunately, in M 12 your code works incorrectly. Easy adjustment solves the problem:

X:= proc(n)

option remember;

     n*(X(n-3)+X(n-2)+X(n-1))

end proc:

X(0),X(1),X(2):= 0,1,2:

length(X(2013));

              5782

@Suji  '$'(1,6)  means the sequence  1,1,1,1,1,1 . Another variant is  1 $ 6 .

@Markiyan Hirnyk  Thanks for the hint!

Corrected code:

Fourier := proc (f, n::{infinity, nonnegint})

local a0, a, b;

a0 := (1/2)*(int(f(x), x = -Pi .. Pi))/Pi;

a := proc (k) options operator, arrow; `assuming`([int(f(x)*cos(k*x), x = -Pi .. Pi, AllSolutions)], [k::posint])/Pi end proc;

b := proc (k) options operator, arrow; `assuming`([int(f(x)*sin(k*x), x = -Pi .. Pi, AllSolutions)], [k::posint])/Pi end proc; 

if n = infinity then return a0+Sum(simplify(a(k)*cos(k*x)+b(k)*sin(k*x), assume = k::posint), k = 1 .. n) else

a0+add(a(k)*cos(k*x)+b(k)*sin(k*x), k = 1 .. n) end if;

end proc;

Now there is no the error.

@Markiyan Hirnyk  This is the bug in Maple. See

int(sin(x)*sin(k*x), x = -Pi .. Pi) assuming k::posint;

                                         0

The result is wrong for k = 1

@Carl Love  In  M 12 only  sort  without options does not work:

sort([[2,1],[3,2],[5,3],[1,4],[4,5]]);

                  [[2, 1], [3, 2], [5, 3], [1, 4], [4, 5]]

@ANANDMUNAGALA   Jacobi's method requires only the simplest of the rotation matriсes, such as those

See the solution http://www.mapleprimes.com/questions/200181-What-Is-The--Maple-Procedure-To-Find

@brian bovril  We must prove that outside a bounded interval there are no roots. It is not enough to refer to the plot of the function.

@Carl Love  Thanks a lot for the solution of the problem!

@Carl Love In Maple 12 (Classic) and Maple 16 (Standard) the result is correct. May be this is a bug in Maple 17?